Taylor's Theorem for Complex Analytic Functions #
Formalization of Theorem 1 (Taylor's Theorem) from Lecture 10 of MIT 18.112 (Functions of a Complex Variable, Fall 2008).
Main results #
complexTaylorPoly— the complex Taylor polynomial of degreen-1.taylor_complex— Part 1: Iffis analytic in a regionΩcontaininga, thenf(z) = f(a) + f'(a)/1! (z-a) + ⋯ + f^(n-1)(a)/(n-1)! (z-a)^(n-1) + fₙ(z)(z-a)^nwherefₙis analytic (differentiable) inΩ.taylor_integral_representation— Part 2: The integral representation of the remainder,fₙ(z) = (2πi)⁻¹ ∮_C f(ζ) dζ / ((ζ-a)^n(ζ-z)), proved by induction onnusing Cauchy's integral formula and a vanishing integral lemma.
Strategy #
The remainder function is defined as the n-fold iterated divided slope (swap dslope a)^[n] f.
The key ingredients are:
dslope_expansion_identity— algebraic identity relatingfto iterated dslope values.iterate_dslope_differentiableOn— iterated dslope preserves differentiability on open sets, usingdifferentiableOn_dslopefrom the removable singularity theory.iterate_dslope_eq_iteratedDeriv_div_factorial— connects dslope coefficients toiteratedDeriv k f a / k!via formal power series machinery.
The n-fold iterated dslope preserves differentiability on open sets.
Uses the complex removable singularity theorem (differentiableOn_dslope).
The coefficient identity: if f has a formal power series expansion at a, then
the value of the k-th iterated dslope at a equals f^(k)(a) / k!.
Taylor's Theorem (Lecture 10, Theorem 1, Part 1).
If f is analytic (holomorphic) in a region Ω containing a, then for every n,
$$f(z) = f(a) + \frac{f'(a)}{1!}(z-a) + \cdots + \frac{f^{(n-1)}(a)}{(n-1)!}(z-a)^{n-1}
- f_n(z)(z-a)^n,$$
where
fₙis analytic (differentiable) inΩ.
Vanishing integral: ∮ 1/((ζ-a)^(n+1) * (ζ-z)) dζ = 0 for z ∈ ball(a, R) and n ≥ 0.
Proved by induction on n using partial fractions and the residue computation
∮ (ζ-w)^m dζ = 0 for m ≠ -1.
Integral representation of the Taylor remainder (Lecture 10, Theorem 1, Part 2).
If C is the boundary of a closed disk of radius R centered at a contained in Ω, then
the remainder function fₙ satisfies
$$f_n(z) = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)\,d\zeta}{(\zeta - a)^n (\zeta - z)}$$
for z inside C.
The proof uses Cauchy's integral formula and induction on n. The key auxiliary result is
circleIntegral_inv_pow_sub_mul_sub_eq_zero, a vanishing integral proved by partial fractions.